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Suppose that I have an $S,V,N$ ensemble. Every variable is a function of the other variable: $U(S,V,N)$, $S(U,V,N)$, $V(S,U,N)$ and $N(S,U,V)$. The functions are everywhere differentiable. But there should also be extreme points. Consider the maximum of $S$. Here we have $(\frac{dS}{dU})_{V,N}=0$. What is now the value of $(\frac{dU}{dS})_{V,N}$? I would expect by the chain rule that $1=(\frac{dS}{dU})_{V,N}\cdot (\frac{dU}{dS})_{V,N}=0\cdot (\frac{dU}{dS})_{V,N}=0$, but that is a contradiction. What am I doing wrong?

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    $\begingroup$ See e.g. here. It has nothing to do with thermodynamics per se. $\endgroup$ Commented Sep 12, 2023 at 11:28
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    $\begingroup$ There might be confusion about which of them are state functions, and which are variables. Some can serve as both, but not simultaneously - the transition requires a Legendre transformations. See Thermodynamic potential. Specifically, in $S, V, N$ ensemble $S$ is not a function of state, but an independent variable. $\endgroup$
    – Roger V.
    Commented Sep 12, 2023 at 11:35
  • $\begingroup$ So in this ensemble, $(\frac{dS}{dU})_{V,N}$ is not defined? What would be a potential for which we can take the derivative of $S$ with respect to that potential? $\endgroup$
    – Riemann
    Commented Sep 12, 2023 at 11:46
  • $\begingroup$ Also an exercise of my physics course says: 'by now you are familiar with the formula $$dU=TdS-PdV+\mu dN.$$ Rewrite this equation to the form $dS=...$, and find expressions for $(\frac{dS}{dV})_{U,N}$ and $(\frac{dS}{dN})_{U,V}$.' How can this be possible if $S$ is not a function of state? $\endgroup$
    – Riemann
    Commented Sep 12, 2023 at 11:50
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    $\begingroup$ when you have $1=(\frac{dS}{dU})_{V,N}\cdot (\frac{dU}{dS})_{V,N}=0\cdot (\frac{dU}{dS})_{V,N}=0$ it is, probably, because $(\frac{dU}{dS})_{V,N}=\infty$. And then you are calculating the product $0 \cdot \infty $ that can be anything. $\endgroup$
    – hyportnex
    Commented Sep 12, 2023 at 15:51

2 Answers 2

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The concepts 'entropy is maximized' and 'energy is minimized' have a different meaning than I thought. The entropy is only maximized with respect to $V,N$ but not with respect to $U$. The energy is only minimized with respect to certain variables and not with respect to $S$. The value of $(\frac{\delta U}{\delta S})_{V,N}$ equals $T$, which is positive. The value of $(\frac{\delta S}{\delta U})_{V,N}$ equals $\frac1T$, which is also positive.

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I think there is a confusion about the second law of thermodynamics. The first thing you need to do is pick an ensemble. This means you need to pick what variables you will arbitrarily hold fixed, and which variables you allow to relax. If you pick the $S$, $V$, $N$ ensemble, you cannot write the entropy as a function of the other variables; you already control the entropy independently!

The clearest ensemble to understand the second law is the $UVN$ ensemble. Here, the thermodynamic potential is the entropy $S(U, V, N)$. The entropy is not maximized with respect to $U$, $V$ or $N$. It is maximized with respect to all unconstrained variables, most of which we do not even think about when we are doing equilibrium statistical mechanics. The change of entropy when you adjust the state variables (like $U$, $V$, $N$) gives rise to thermodynamic forces, like temperature, pressure and chemical potential. You can think of the state variables as constraints you impose on the system.

Now, when you calculate $\partial S / \partial U$, you calculate the thermodynamic force that 'pulls on you' as you increase the internal energy. This is (one over) the temperature of the system: $\partial S / \partial U = 1/T$. Similarly, when you calculate $\partial S / \partial V = p/T$, you calculate the pressure of the system.

The second law of thermodynamics says that if you relax the constraints you put on the system a little, the system can and will go to a state of higher entropy. The derivatives of the entropy tell you how much of an entropy gain the system could achieve by working against your constraints, which corresponds to the 'force' the system exerts against you.

Regarding your question what happens at points where $\partial S / \partial U = 1/T = 0$: The inverse function rule tells you that at that point $\partial U / \partial S = (\partial S / \partial U)^{-1} = \infty$. From a more physical perspective, this point corresponds to infinite temperature.

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  • $\begingroup$ What would be some of those unconstrained variables? $\endgroup$
    – Riemann
    Commented Sep 19, 2023 at 10:18

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